305. Number of Islands II

A 2d grid map of m rows and n columns is initially filled with water. We may perform an addLand operation which turns the water at position (row, col) into a land. Given a list of positions to operate, count the number of islands after each addLand operation. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.

Example:

Given m = 3, n = 3, positions = [[0,0], [0,1], [1,2], [2,1]].
Initially, the 2d grid grid is filled with water. (Assume 0 represents water and 1 represents land).

0 0 0
0 0 0
0 0 0

Operation #1: addLand(0, 0) turns the water at grid[0][0] into a land.

1 0 0
0 0 0   Number of islands = 1
0 0 0

Operation #2: addLand(0, 1) turns the water at grid[0][1] into a land.

1 1 0
0 0 0   Number of islands = 1
0 0 0

Operation #3: addLand(1, 2) turns the water at grid[1][2] into a land.

1 1 0
0 0 1   Number of islands = 2
0 0 0

Operation #4: addLand(2, 1) turns the water at grid[2][1] into a land.

1 1 0
0 0 1   Number of islands = 3
0 1 0

We return the result as an array: [1, 1, 2, 3]

Challenge:

Can you do it in time complexity O(k log mn), where k is the length of the positions?

Similar:

• 300. Number of Islands

public class Solution {
    public List<Integer> numIslands2(int m, int n, int[][] positions) {
        List<Integer> cnt = new ArrayList<Integer>();
        int[][] root = new int[m][n];
        for (int i=0; i<m; i++)
            for (int j=0; j<n; j++) {
                root[i][j] = -1;
            }
        int lastNumber = 0;
        
        for (int[] pt : positions) {
            lastNumber = addLand(root, pt[0], pt[1], lastNumber);
            cnt.add(lastNumber);
        }
        
        return cnt;
    }
    
    private int addLand(int[][] root, int r, int c, int num) {
        if (root[r][c] != -1) return -1; // used to land this part
        int m = root.length;
        int n = root[0].length;
        
        root[r][c] = r * n + c; // root is itself
        num += 1;
        
        if (r-1>=0 && root[r-1][c]!=-1) { // up
            if (union(root, r-1, c, r, c))
                num--;
        }
        if (r+1<m && root[r+1][c]!=-1) { // down
            if (union(root, r+1, c, r, c))
                num--;
        }
        if (c-1>=0 && root[r][c-1]!=-1) { // left
            if (union(root, r, c-1, r, c))
                num--;
        }
        if (c+1<n && root[r][c+1]!=-1) { // right
            if (union(root, r, c+1, r, c))
                num--;
        }
        return num;
    }
    
    private boolean union(int[][] grid, int r1, int c1, int r2, int c2) {
        int root1 = find(grid, r1, c1);
        int root2 = find(grid, r2, c2);
        
        if (root1 == root2) return false; // already in one set
        
        int n = grid[0].length;
        int x = root2 / n;
        int y = root2 % n;
        
        grid[x][y] = root1;
        return true;
    }
    
    private int find(int[][] grid, int r, int c) {
        int n = grid[0].length;
        int id = r * n + c;
        int rr, cc;
        while (id != grid[r][c]) {
            rr = grid[r][c] / n;
            cc = grid[r][c] % n;
            grid[r][c] = grid[rr][cc];
            r = rr;
            c = cc;
            id = rr * n +cc;
        }
        return id;
    }
}